1. **Problem Statement:** Given a technology matrix $A$ and a production schedule vector $X$, find the external demand vector $D$ such that $$X = AX + D.$$ 2. **Formula and Explanation:** Rearranging the equation, we get $$D = X - AX.$$ This means the external demand $D$ is the difference between the total production $X$ and the internal consumption $AX$. 3. **Step 1: Define the technology matrix $A$** From Examples 6.17-6.18, assume the technology matrix is: $$A = \begin{bmatrix}0.3 & 0.1 \\ 0.2 & 0.4\end{bmatrix}$$ 4. **Step 2: Calculate $D$ for $X = \begin{bmatrix}30 \\ 70\end{bmatrix}$** Calculate $AX$: $$AX = \begin{bmatrix}0.3 & 0.1 \\ 0.2 & 0.4\end{bmatrix} \begin{bmatrix}30 \\ 70\end{bmatrix} = \begin{bmatrix}0.3 \times 30 + 0.1 \times 70 \\ 0.2 \times 30 + 0.4 \times 70\end{bmatrix} = \begin{bmatrix}9 + 7 \\ 6 + 28\end{bmatrix} = \begin{bmatrix}16 \\ 34\end{bmatrix}$$ Then, $$D = X - AX = \begin{bmatrix}30 \\ 70\end{bmatrix} - \begin{bmatrix}16 \\ 34\end{bmatrix} = \begin{bmatrix}14 \\ 36\end{bmatrix}$$ 5. **Step 3: Calculate $D$ for $X = \begin{bmatrix}120 \\ 330\end{bmatrix}$** Calculate $AX$: $$AX = \begin{bmatrix}0.3 & 0.1 \\ 0.2 & 0.4\end{bmatrix} \begin{bmatrix}120 \\ 330\end{bmatrix} = \begin{bmatrix}0.3 \times 120 + 0.1 \times 330 \\ 0.2 \times 120 + 0.4 \times 330\end{bmatrix} = \begin{bmatrix}36 + 33 \\ 24 + 132\end{bmatrix} = \begin{bmatrix}69 \\ 156\end{bmatrix}$$ Then, $$D = X - AX = \begin{bmatrix}120 \\ 330\end{bmatrix} - \begin{bmatrix}69 \\ 156\end{bmatrix} = \begin{bmatrix}51 \\ 174\end{bmatrix}$$ **Final answers:** - For $X = \begin{bmatrix}30 \\ 70\end{bmatrix}$, external demand $D = \begin{bmatrix}14 \\ 36\end{bmatrix}$. - For $X = \begin{bmatrix}120 \\ 330\end{bmatrix}$, external demand $D = \begin{bmatrix}51 \\ 174\end{bmatrix}$. These results show how much external demand is needed to satisfy the given production schedules after accounting for internal consumption.